Different type of air standard ( otto and diesel cycle)

7/29/2019 Different type of air standard ( otto and diesel cycle)

1/23

Air Standard Cycle

The accurate analysis of IC process is very completed. In order to

Because of the complex chemical reactions that take place when

fuel burns and friction and heat transfer between the gases and

the cylinder. It is therefore a usual practice to analyze the cycleMaking some simplifying assumptions.

The assumptions are as follows:

The working fluid is a perfect/ideal gas through out i.e. it

follows the PV= mRTm = mass (Kg), V= volume occupied(m3),

= Universal gas constant (8.314 KJ/Kg oK)

M = Molecular weight of gas

M

R__

M

R__

M

RR

__

__R

Engr. Md. Irteza Hossain, Faculty,Mechanical Engineering, IUBAT


2/23

Air Standard Cycle

The working fluid is a fixed mass of air either contained in a closed

system or flowing at a constant rate round a closed circuit. The intake

and exhaust process are not considered.

The physical constants of the working medium are the same as those

of air at standard atmospheric conditions, i.e molecular weight=29

Specific heat at constant pressure (CP)= 1.005 KJ/kgoK

Specific heat at constant volume ( Cv)= 0.718 KJ/kgo

K

Adiabatic index, 4.1718.0

005.1

CV

Cp



3/23

Air Standard Cycle

The working medium has a constant specific heat having following

relation:

Cp- Cv = R and CV=

The working medium does not undergo any chemical change throughout the cycle

The combustion process is replaced by a heat transfer process from

an external source.

The cycle is completed by heat rejection to the surrounding until theair temperature corresponds to initial condition

All process that constitute the cycle are reversible i.e no loss or gain

of entropy

1)(

R



4/23

Air Standard Cycle

The compression and expansion processes are reversible adiabatic

The operation of the engine is frictionless.

Because of the simplified assumptions, the peak temperature, the

pressure, the work output and the thermal efficiency calculated by the

Analysis of an air standard cycle are higher than those found in anActual engine. However, the analysis shows the relative effects of the

principal variable such as compression ratio, inlet pressure, inlet

temperature effect on the engine performance

The main use of ideal cycle is that it is amenable to simple

mathematical calculation based on perfect gas law and can give

rough answer on the effect on principal values such as r, inlet t, P



5/23

Otto Cycle or constant volume cycle

Process 1-2 : Reversible adiabaticor isentropic process. There is no

heat transfer

Process 2-3: Reversible constantvolume heating

Process 3-4: Reversible adiabaticor isentropic expansion. There is no

heat transfer

Process 4-1: Reversible constantvolume heat rejection.



6/23

Otto Cycle or constant volume cycleDerivation of Thermal

efficiency and mean effectivepressure:

Vs = Swept volume

Vc = Clearance volume

V1= total cylinder volume = Vs + Vc

Compression ration

SuppliedHeat

rejectedHeatSuppliedHeat

SuppliedHeat

doneWorkEf f iciencyThermalThe

V

V

Vc

VcVs

volumeClearnce

volumecylinderTotalrrationCompressio

,

2

1,



7/23

Thermal Efficiency of Otto Cycle

--------------------------------------------------------------

---------------------------------------------------------------

1

2

1pp

1Q

Q

liedsuHeat

rejectedHeat

r1

1

1



8/23

The effect of compression ratio on thermal efficiency

The thermal efficiency of

the cycle increase with increaseof compression ratio

At a higher value of adiabatic

exponent thermal

efficiency() is also increasesFor a given value of

compression ratio and adiabatic

exponent , the thermal value is

constant and does not depend onamount heat supplied

Heat supplied is directly

proportional to the load and hence

the thermal efficiency does not

depend on the engine load



9/23

Mean Effective Pressure

Mean effective Pressure,

VolumeSwept

DoneWorkPpressureEffectiveMean m )(

Vs

W

diagramtheofLength

diagramindicatortheofArea

r

PPpressureEffectiveMean rm

)1)(1(

)1()(

1

additionheatatratiopressure

barpressureInitialpasWhere

)(1



10/23

Mean effective pressure

)1)(1(1

1

rTmcW v

The work output of the cycle increases with pressure

ratio() and compression ratio (r) The increase of the pressure ratio result in increase of

amount of heat received



11/23

ANALYSIS

Variation of Pm versus Pressure ratio()

Pm increase with Pressure ratio() and

compression ratio (r)

All the analysis of Pm and of the cycleshows that using higher compression

ratio (r) is the effective way of improving

the performance of the engine

> the value of r is restricted to ensure of avoiding auto ignitionand detonation

)1)(1(

)1()(

1

r




12/23

Problem and solution A gas engine working on the otto cycle has cylinder bore of

200mm and stroke length of 250 mm. The clearance volume is

1570 cm3. The pressure and temperature at the beginning ofcompression are 1 bar and 27 oC respectively. The maximumtemperature of the cycle is 1400 oC. Determine the pressure andtemperature at the salient points, the air standard efficiency,the work done and mean effective pressure . ( For air , take Cv=

0.718 KJ/Kgo

K) and R =0.287 KJ/Kgo

K).Also calculate the idealpower developed by the engine, if the number of workingcycles per minute is 500.



13/23

Problem and solution

In an ideal otto cycle the air at the beginning of isentropic

compression is at 1 bar at 15oC. The compression ratio is 8. Ifthe heat rejected during the constant volume process is 1000KJ/Kg, determine

The maximum temperature in the cycle

The air standard cycle efficiency

The work done per kg of air

The heat rejected

Considering Cv=0.718 and =1.4



14/23

Problem and solution In a petrol engine the swept volume ( Stroke volume) is 0.13 m3.

The temperature T1= 2000

o

K, T2= 977

o

K, T3= 333

o

K and T4= 681oK. The engine produces power stroke one in 2 revolution. The

Engine runs at 1000 rpm. Calculate

Heat supplied

Heat rejected Work done ( all per cycle and per minute)

If CV of fuel is 4500 J/Kg, what amount of fuel is required per minute?

Use Cv = 713 J/Kg oK , m= 0.1615 Kg.



15/23

Diesel Cycle or constant pressure cycle In SI engine the upper limit of compression ratio is

limited by self ignition temperature of the fuel This limitation of compression ration is circumvented if

air and fuel is compressed separately and bought

together at the time of combustion

The fuel is injected in to the cylinder which containscompressed air at a high temperature than self ignition

temperature of the fuel

The fuel is ignite on its own accord and requires no

special device for ignition



16/23

Diesel Cycle or constant pressure cycle

Process 1-2 : Reversible adiabaticor isentropic process. There is noheat transfer

Process 2-3: Reversible constantpressure heating

Process 3-4: Reversible adiabaticor isentropic expansion. There is no

heat transfer

Process 4-1: Reversible constantvolume heat rejection.



17/23

Diesel Cycle or constant pressure cycle

Derivation of Thermalefficiency and mean effectivepressure:

Vs = Swept volume

Vc = Clearance volume

V1= total cylinder volume = Vs + Vc

Compression ration

SuppliedHeat

rejectedHeatSuppliedHeat

SuppliedHeat

doneWorkEff iciencyThermalThe

V

V

Vc

VcVs

volumeClearnce

volumecylinderTotalrrationCompressio

,

2

1,


)(

)(int)(

2

3

VvolumeClearance

VcutoffofpotheatVolumeratiooffCut


18/23

Thermal Efficiency of Diesel Cycle

--------------------------------------------------------------

---------------------------------------------------------------

1

2

1pp

1Q

Q

liedsuHeat

rejectedHeat

])1(

1

[

1

1 1

r



19/23

Mean Effective Pressure

Mean effective Pressure,

Pm increases with initial pressure p1, adiabatic index , r. The cut off

ratio increases when large amount of heat received, which increases Pm,

and reduces

VolumeSwept

DoneWorkPpressureEffectiveMean m )(

Vs

W

diagramtheofLengthdiagramindicatortheofArea

r


)1)(1(

)1()(

1


)()( 142321 TTmcTTmcQQW vp


20/23

Diesel Cycle or constant pressure cycle The thermal efficiency of a diesel cycle depends on the

compression ration r, adiabatic exponent & the cut offratio .

The air standard thermal efficiency of otto cycle is

The air standard thermal efficiency of Diesel cycle

Diesel cycle differs from otto cycle by terms which

is always greater than unity. Therefore the air standard

efficiency of diesel cycle is less than otto cycle for same

compression ratio.Engr. Md. Irteza Hossain, Faculty,Mechanical Engineering, IUBAT

r1

11

])1(

1[

11

1

r]

)1(

1[


21/23

Graphical Analysis

The following figure shows that the thermal efficiency of a Diesel

cycle depends on the compression ratio( r), the adiabatic index For a given value of the cut- off ration , the efficiency of the diesel

cycle increases with increase of cut off ratio



22/23

Thermal efficiency



23/23

Problem and solution

An ideal diesel engine has a diameter of 15 cm and stroke 20

cm. The clearance volume is 10 percent of the swept volume.Determine the compression ratio and the air standardefficiency of the engine if the cut- off takes place at 6 percentof the stroke.

In an engine working on Diesel cycle, the air fuel is 30:1. The

temp of the air at the beginning of compression is 27oC., thecompression ratio is 16:1. What is the ideal efficiency of theengine? The calorific value of the fuel is 42000Kj/Kg

In a engine working on Diesel cycle, the inlet pressure is 1 bar.The pressure at the end of isentropic compression is 32.425

bar. The ratio of expansion is 6. Calculate the air standardefficiency and the mean effective pressure of the cycle.


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Different type of air standard ( otto and diesel cycle)